Obviously troubles

Since obviously learned a tree, strange tree on the right became interested in ...... 

Reference numeral 1 is given to the point N, and the final degree certain points, allowing the connection between any two points, the number of degrees trees tree can be generated to meet the requirements?
The Input
first row N (0 <N <= 1000 )
Next N rows, row i + 1-Di given degree of node i, if the degree is not required, enter -1
output
an integer representing the number to meet the requirements of different trees, no solution output 0
the Sample INPUT
. 3
. 1
-1
-1
the Sample the Output
2
the HINT
two trees, respectively 1-2-3; 1-3-2 

input 

8
2
2
-1
-1
-1
-1
-1
-1

output

38880

Sol:
Purfer 8-2 total position coding
found in 1 of 2, 1 appears, then C (6,1)
found of 2 is 2, 1 appears, then C (5,1)
as well as 4 position is not put things, so that no limitation of the discharge point 6, 6 ^ 4 ways
so ans = 38880

 1 #include<queue>
 2 #include<cmath>
 3 #include<cstdio>
 4 #include<vector>
 5 #include<cstring>
 6 #include<iostream>
 7 #include<algorithm>
 8 #define fi(i,a,b) for(int i=(a);i<=(b);++i)
 9 #define fj(j,a,b) for(int j=(a);j<=(b);++j)
10 #define maxn 1005
11 #define inf 1000000000
12 using namespace std;
13 inline long long read()
14 {
15     long long x=0,f=1;char ch=getchar();
16     while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
17     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
18     return x*f;
19 }
20 int n,cnt,sum;
21 int d[maxn],prime[maxn],c[maxn],t[maxn*20];
22 bool flg,p[maxn];
23 void getprime(int x)
24 {
25     memset(p,1,sizeof(p));
26     prime[0]=1; prime[1]=2;
27     fi(i,2,x)if(p[i]){fj(j,2,x/i)p[i*j]=0;prime[++prime[0]]=i;}
28 }
29 void update(int x,int m)
30 {
31     for(int i=1;i<=prime[0]&&x>1;++i)
32     while(!(x%prime[i]))
33     {
34         x/=prime[i];c[i]+=m;
35     }
36 }
37 int main()
38 {
39 //  freopen("data.in","r",stdin);
40     n=read(); getprime(1000);
41 //  fi(i,1,prime[0])printf("%d\n",prime[i]);
42     fi(i,1,n){
43         d[i]=read();
44         if(!d[i]||d[i]>=n)flg=1;
45         if(d[i]==-1)continue;
46         ++cnt; sum+=d[i]-1;
47         fj(j,2,d[i]-1)update(j,-1);
48     }
49     if(n==1&&d[1]>0){printf("1");return 0;}else if(n==1)flg=1;
50     if(n==2&&!flg){printf("1");return 0;}
51     if(flg){printf("0\n");return 0;}
52     fi(i,2,n-2)update(i,1);
53     fi(i,2,n-2-sum)update(i,-1);
54     update(n-cnt,n-2-sum);
55     t[0]=t[1]=1;
56 //  printf("%d\n",c[1]);
57     fi(i,1,prime[0])while(c[i]){
58         if(c[i]<0){printf("0\n");return 0;}
59 //      printf("%d %d\n",prime[i],c[i]);
60 //      printf("%d\n",prime[i]);
61         --c[i];
62         fj(j,1,t[0])t[j]*=prime[i];
63         fj(j,1,t[0]-1)t[j+1]+=t[j]/10,t[j]%=10;
64         while(t[t[0]]>=10)
65         {
66             t[t[0]+1]+=t[t[0]]/10;
67             t[t[0]]%=10;
68             ++t[0];
69         }
70     }
71     for(int i=t[0];i;--i)printf("%d",t[i]); printf("\n");
72 }